Equilibrium Yield Graphs

UNIT 2 CHEMISTRY · PEARSON AAQ · D1.4.5

Equilibrium Yield Graphs

FREE CHEMISTRY REVISION · TOPIC 47

Equilibrium Yield Graphs

Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.4.5

What you will learn

Read equilibrium yield against pressure and temperature, identify the controlled conditions, and explain each trend using gas mole ratios and reaction enthalpy. These are equilibrium composition graphs: their gradients are not reaction rates.

1. Read the axes and conditions first

Equilibrium yield here means the amount of the named product present at equilibrium divided by its maximum possible amount from the same starting feed, multiplied by 100%. It is not necessarily the product’s mole percentage in the mixture. Every plotted point represents a mixture allowed to reach equilibrium.

Identify the product named on the vertical axis and its percentage scale.
Read the horizontal variable and units: total pressure in atm or temperature in K.
Identify which curve fixes temperature or pressure. Keep the starting feed the same.
Move vertically from the chosen horizontal value to the curve, then horizontally to read yield.
Use approximate readings consistent with the graph resolution.
Describe the change, then explain it using the balanced equation.

The pressure graphs below mean compression or expansion of the same closed gas mixture at constant temperature. Adding an inert gas at constant volume raises total pressure without changing the reacting gases’ partial pressures, so it need not change yield. Always identify how pressure was changed.

2. Pressure: count gaseous coefficients

For 2NO₂(g) ⇌ N₂O₄(g), the forward reaction is exothermic and changes two moles of gas to one. Compression increases pressure; the equilibrium shifts towards fewer gas moles, increasing the dimer N₂O₄ yield. Expansion favours the two-mole NO₂ side.

Graph A. N₂O₄ equilibrium yield from a fixed feed equivalent to 2 mol NO₂. Each curve has a constant temperature. Lower temperature favours the exothermic dimer-forming direction. These original curves are an illustrative ideal-gas model, not experimental measurements.

Worked interpretation 1 — pressure increase

Follow the 298 K curve in Graph A. At 1 atm, N₂O₄ yield is about 81%; at 4 atm, about 90%. The increase is approximately 9 percentage points. Pressure favours the one-gas-mole product side over the two-gas-mole reactant side.

Do not call this a 9% relative increase. Using the model values, (90.4 − 81.0)/81.0 × 100 ≈ 12% relative increase. A graph reading normally warrants whole percentages, not several decimal places.

Worked interpretation 2 — compare temperature curves

At the same pressure of 1 atm, Graph A gives about 81% N₂O₄ at 298 K and 49% at 330 K. Raising temperature decreases dimer yield by about 32 percentage points. Heating favours the endothermic reverse dissociation. It does not follow that the reaction is slower at 330 K: rate and equilibrium yield answer different questions.

3. Temperature: identify the endothermic direction

Reverse the equation: N₂O₄(g) ⇌ 2NO₂(g). The forward dissociation is endothermic, with positive ΔH. Increasing temperature favours NO₂ formation; decreasing temperature favours N₂O₄. The sign of ΔH belongs to the reaction as written, so reversing the equation reverses the sign.

Graph B. NO₂ equilibrium yield from 1 mol initial N₂O₄, relative to the maximum 2 mol NO₂. Each curve fixes total pressure. The lower-pressure curve has greater NO₂ yield because this product side has more gas moles.

Worked interpretation 3 — temperature increase

Follow the 1 atm curve in Graph B. At 298 K the NO₂ yield is about 19%; at 330 K it is about 51%. The rise is about 32 percentage points. Heating favours the endothermic dissociation direction. The temperature axis is in kelvin, not degrees Celsius.

Worked interpretation 4 — pressure at a fixed temperature

At 330 K, compare the 0.5 atm and 2 atm curves in Graph B: NO₂ yields are about 64% and 39%, respectively. Compression lowers NO₂ yield by roughly 25 percentage points, because it favours the one-mole N₂O₄ side. The low-pressure curve is therefore above the high-pressure curve.

Graph A and Graph B name different products. At the same temperature and pressure in this model, their percentage yields sum to 100: at 330 K and 1 atm, about 49% dimer plus 51% dissociation. This complement follows from the specified starting feed and yield definitions; it is not a rule for every reaction.

4. Equal gas mole totals: a flat pressure graph

H₂(g) + I₂(g) ⇌ 2HI(g) has two gas moles on each side. At a fixed temperature, uniformly changing the gas partial pressures by compression or expansion leaves their ratio p(HI)²/(p(H₂)p(I₂)) unchanged in the ideal-gas model. There is no equilibrium shift.

Graph C. An illustrative fixed-temperature model with Kp = 16 and initially equal amounts of H₂ and I₂, with no HI. Iodine is gaseous under the stated model conditions. All pressures shown give the same equilibrium HI yield, approximately 67%.

Worked interpretation 5 — explain a horizontal line

Read Graph C at 1 atm and 4 atm: both give about 67% HI yield. Equal gaseous coefficient totals explain the absence of a pressure effect. A flat yield curve does not mean the reaction has stopped or has zero rate: forward and reverse reactions continue at equal rates at equilibrium.

Optional check of the plotted level: start with 1 mol each of H₂ and I₂. If x mol of each reacts, HI formed = 2x and each reactant remaining = 1 − x. Kp = 4x²/(1 − x)² = 16 gives 2x/(1 − x) = 4, hence x = 2/3. Yield = 2x/2 × 100 = 66.7%. Every amount is non-negative.

5. Catalysts, rate and equilibrium yield

Worked interpretation 6 — adding a catalyst

A suitable catalyst is introduced to a system represented by Graph C, at the same temperature, pressure and starting composition. The equilibrium point remains on the same approximately 67% line. A catalyst speeds both forward and reverse reactions and allows equilibrium to be reached sooner; it does not change K or the equilibrium yield.

If a measurement made too soon shows more product with the catalyst, that can be a rate effect before equilibrium. It is not evidence that the equilibrium curve moved. No particular catalyst or practical preparation is specified by this illustrative comparison.

Observation	Equilibrium interpretation	What it does not establish
Yield rises with compression	Products have fewer gas moles, for the stated ideal-gas conditions	The reaction necessarily becomes faster by the same proportion
Yield rises with temperature	Product-forming direction is endothermic	The yield-time curve has the same shape
Yield is unchanged by pressure	Equal gas mole totals can explain this	Pressure never affects any equilibrium
Catalyst gives the same final yield sooner	Equilibrium position is unchanged	The catalyst increases K

6. How the original graphs were constructed

Graphs A and B use one internally consistent ideal-gas teaching model. For N₂O₄ ⇌ 2NO₂, the illustrative dissociation Kp is 0.150 atm at 298 K and the assumed constant ΔH is +57.0 kJ mol⁻¹. Values at other temperatures follow ln(Kp(T)/Kp(298)) = −ΔH/R × (1/T − 1/298), with R = 8.314 J mol⁻¹ K⁻¹. These assumptions generate teaching curves, not a claim of measured accuracy.

With 1 mol initial N₂O₄ and dissociation fraction α, equilibrium amounts are 1 − α and 2α mol; total amount is 1 + α. Therefore Kp = 4α²P/(1 − α²) and α = √(Kp/(Kp + 4P)). Graph B plots 100α; Graph A plots 100(1 − α). All plotted yields stay between 0% and 100%. No inert gas is present. These model details are provided for transparency; interpreting the graphs does not require deriving this formula.

Common misconceptions

Higher pressure always increases yield: it depends on which side has fewer gaseous moles; equal totals can give no change.
Count solids in the gas-mole ratio: count gaseous coefficients only.
Higher temperature always gives more product: identify the endothermic direction first.
A steep equilibrium curve means a fast reaction: the horizontal axis is not time.
A catalyst increases equilibrium yield: it changes the time to equilibrium, not the equilibrium position.
Yield percentage is mole percentage: check the defined maximum product amount.
A smooth curve justifies many decimal places: give readings consistent with scale and line thickness.

Examination tips

Quote both conditions when reading a family of curves.
Use approximate values and calculate percentage-point changes correctly.
Link pressure explanations to the balanced gaseous mole ratio.
Link temperature explanations to the sign of ΔH for the stated forward reaction.
Say equilibrium shifts, not that only one reaction occurs.
Keep conclusions about equilibrium composition separate from conclusions about reaction rate.

Check your understanding

Choose one answer, select Check Answer, then read the explanation. Try Again resets the question. Use Tab between controls, arrow keys within a radio group, and Enter or Space on buttons. Answers are not saved.